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So I am reading this paper https://arxiv.org/pdf/math/0008177.pdf by Jeffrey Lagarias and in the proof of Lemma 3.1 he says \begin{equation} \int_{1}^{n} \frac{\lfloor t \rfloor}{t^2}dt = \sum_{1 \le r \le n} \int_{r}^{n} \frac{1}{t^2} dt. \end{equation} In the paper he uses $\infty$ as the upper integration bound, but I think it should be n, and r as the lower end, in order for the next steps in the paper to make sense. The left side is equal to $\int_{1}^{n} \sum_{1 \le r \le t} \frac{1}{t^2} dt$, but I cannot see how \begin{equation} \int_{1}^{n} \sum_{1 \le r \le t} \frac{1}{t^2} dt = \sum_{1 \le r \le n} \int_{r}^{n} \frac{1}{t^2} dt \end{equation} would hold. Yes it is a finite sum, but we are integrating over t, so we shouldn't be able to just move it out of the sum, right?

Then I thought it might help that $\lfloor t \rfloor$ is a step function, but we multiply it with $\frac{1}{t^2}$, so I can't see how I can use that either. I tried using integration by parts but I got a completely different result than whats in the paper. WolframAlpha tells me it exceeds computation time. What am I missing here?

I apologise if this is a stupid question, it's been a long time since my Analysis courses!

Holly Morph
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  • Lagarias' upper limit of $\infty$ is almost surely correct, but the equation is not correct as written with only that change. Did you also change anything else? Be that as it may, the usual way to bridge such a gap is not to try to manipulate one side to the other, but to evaluate each side and see whether they're equal. :) – Andrew D. Hwang Apr 01 '21 at 11:56
  • @AndrewD.Hwang: I cannot find an error in the equation, what am I overlooking? – Martin R Apr 01 '21 at 12:12
  • @Martin: When I calculate each side of the first equation (after setting the upper limit of integration on the right to $\infty$, as OP suggested was in the original), the right side is $1$ larger than the left. But on rereading, it's not unlikely that I misinterpreted what was said to be in Lagarias' paper. – Andrew D. Hwang Apr 01 '21 at 12:28
  • @AndrewD.Hwang: Actually that seems to be an error in that article. $\int_{1}^{\infty} \frac{\lfloor t \rfloor}{t^2}dt = \cdots = H_n - 1$ makes no sense because the LHS does not depend on $n$. – Martin R Apr 01 '21 at 12:43
  • The trick is to express the step function $\lfloor t\rfloor$ as a sum of shifted Heaviside steps. –  Apr 01 '21 at 15:03

1 Answers1

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You have $$ \int_{1}^{n} \frac{\lfloor t \rfloor}{t^2}dt = \sum_{k=1}^{n-1} k \int_{k}^{k+1} \frac{dt}{t^2} $$ because $\lfloor t \rfloor = k$ on each interval $[k, k+1)$. Now write $k$ as a sum $$ \ldots = \sum_{k=1}^{n-1} \sum_{r=1}^{k} \int_{k}^{k+1} \frac{dt}{t^2} \, . $$ and change the order of summation: $$ \ldots = \sum_{r=1}^{n-1} \sum_{k=r}^{n-1} \int_{k}^{k+1} \frac{dt}{t^2} = \sum_{r=1}^{n-1} \int_{r}^{n} \frac{dt}{t^2} \, . $$ That is your right-hand side because the term with $r=n$ is zero.

It may be instructive to write the transformation explicitly for small values of $n$, e.g. $n=4$: $$ \int_{1}^{4} \frac{\lfloor t \rfloor}{t^2}dt = \int_1^2 \frac{dt}{t^2} + 2 \int_2^3 \frac{dt}{t^2} + 3 \int_3^4 \frac{dt}{t^2} \\ = \left( \int_1^2 \frac{dt}{t^2} + \int_2^3 \frac{dt}{t^2} + \int_3^4 \frac{dt}{t^2}\right) + \left( \int_2^3 \frac{dt}{t^2} + \int_3^4 \frac{dt}{t^2}\right) + \left(\int_3^4 \frac{dt}{t^2}\right) \\ = \int_1^4 \frac{dt}{t^2} + \int_2^4 \frac{dt}{t^2} + \int_3^4 \frac{dt}{t^2} \, . $$

Here is another way to look at it (perhaps more in line with your question “Can we integrate over a summation index?”). A sum can be written as an integral with respect to the counting measure $\mu$ on the integers. Then your equation becomes an application of Fubini's theorem:

$$ \int_{1}^{n} \frac{\lfloor t \rfloor}{t^2}dt = \int_{1}^{n} \sum_{1 \le r \le t} \frac{1}{t^2} dt = \int_{1}^{n} \int_1^t \frac{1}{t^2} \, d\mu(r) \, dt \\ = \int_1^n \int_r^n \frac{1}{t^2} \, dt\, d\mu(r) = \sum_{n=1}^r \int_r^n \frac{1}{t^2} \, dt \, . $$

Martin R
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